Conserved quantities with charge and Killing's vectors

Question

I'm trying to solve the following problem:

A particle with electric charge e moves with 4-velocity $U_{\alpha}$ in a spacetime with metric $g_{αβ}$ in the presence of a vector potential $A_µ$. The equation describing this particle’s motion can be written $$U^{\beta} \nabla_{\beta} U_{\alpha} = e(\nabla_{\alpha} A_{\beta} - \nabla_{\beta} A_{\alpha})U^{\beta} $$.

The spacetime admits a Killing vector field $ξ^α$ such that $$\cal L_{\xi} g_{αβ} = 0,$$ $$\mathcal{L}_{\xi} A_α = 0.$$

Show that the quantity $(U_α + eA_α)ξ^α$ is constant along the worldline of the particle.

With all the derivatives introduced in GR, I'm slightly confused about what constant exactly means in this case. Nevertheless, I tried taking the Lie derivative of the above quantity and used the fact that $\cal L_{\xi} A_{\alpha} = 0$ to arrive at the following:

$$ L_{\vec {U}} (U_{\alpha} + eA_{\alpha})ξ^{\alpha} = \xi^{\alpha} e U^{\gamma} \nabla_{\alpha} A_{\gamma} + U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha} + U^{\gamma} e A_{\alpha} \nabla_{\gamma} \xi^{\alpha}, $$

using $\cal L_{\xi} A_{\alpha} = 0$, I can convert $\xi^{\alpha} \nabla_{\alpha} A_{\gamma} = - A_{\alpha} \nabla_{\gamma} \xi^{\alpha}$ to get

$$L_{\vec {U}} (U_{\alpha} + eA_{\alpha})ξ^{\alpha} = U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha}$$

which is clearly not zero. I haven't used the fact that $\cal L_{\xi} g_{αβ} = 0$ but I don't know how to apply it here. Also, I know this question was treated in Conserved quantities from Killing vectors in the presence of electric charge but there they use the action and we haven't talked much about that in the course, so I think it is not necessarily meant to be solved with it. Any ideas?

Answer 1

What you've written all looks correct, and you were very close to the result. You were along the right lines with needing $\mathcal{L}_{\xi}g_{\mu\nu}=0$. Your remaining term can be written as $$ U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha} = U^{\gamma} U^{\alpha} \nabla_{\gamma} \xi_{\alpha} \, , $$ which comes from metric compatibility. The Lie derivative of the metric is $\mathcal{L}_{\xi}g_{\mu\nu}= 2 \nabla_{(\mu} \xi_{\nu)}$, which vanishes for Killing vector fields. Hopefully with these two equations you can now see why the remaining term is zero.